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Thread: Complex Number

  1. #1
    Senior Member vincisonfire's Avatar
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    Complex Number

    Hi, my question is :
    Let n be a positive integer. Find all real numbers x such that $\displaystyle (x-i)^n + (x+i)^n = 0 $
    I find that there is no such REAL number.
    Does someone find a different answer?
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    Quote Originally Posted by vincisonfire View Post
    Hi, my question is :
    Let n be a positive integer. Find all real numbers x such that $\displaystyle (x-i)^n + (x+i)^n = 0 $
    I find that there is no such REAL number.
    Does someone find a different answer?
    Hint: convert to polars and use De Moivre's Theorem.

    You should find that it reduces down to

    $\displaystyle 2(x^2 + 1)^{\frac{n}{2}}\cos{(nx)} = 0$.

    You should be able to solve this with null factor law.
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    Quote Originally Posted by vincisonfire View Post
    Hi, my question is :
    Let n be a positive integer. Find all real numbers x such that $\displaystyle (x-i)^n + (x+i)^n = 0 $
    I find that there is no such REAL number.
    Does someone find a different answer?
    Well a pretty obvious solution from inspection is that if n is even, then x = 0 is a real number solution. But I'll do the working just to see if any others lurk:
    $\displaystyle (x-i)^n = -(x+i)^n $

    $\displaystyle x-i = (-(x+i)^n)^{\frac{1}{n}} $

    $\displaystyle x-i = (-1)^{\frac{1}{n}} (x+i) $

    $\displaystyle x-i = \frac{(-1)^1}{(-1)^n} \times (x+i) $

    $\displaystyle x-i = \frac{-1}{(-1)^n} \times (x+i) $

    $\displaystyle x-i = \frac{-x}{(-1)^n} - \frac{i}{(-1)^n} $

    Get all x terms on LHS and non-x terms on RHS

    $\displaystyle x+ \frac{x}{(-1)^n} = i- \frac{i}{(-1)^n} $

    $\displaystyle x(1+ \frac{1}{(-1)^n}) = \frac{i(-1)^n-i}{(-1)^n} $

    $\displaystyle x(\frac{(-1)^n+1}{(-1)^n}) = \frac{i(-1)^n-i}{(-1)^n} $

    $\displaystyle x = \frac{i(-1)^n-i}{(-1)^n} \times \frac{(-1)^n}{(-1)^n+1} $

    $\displaystyle x = \frac{i(-1)^n-i}{(-1)^n+1} $

    $\displaystyle x = \frac{i((-1)^n-1)}{(-1)^n+1} $

    Hence $\displaystyle n \neq 2k+1 $ for $\displaystyle k$ any positive integer.

    For $\displaystyle n = 2k$ $\displaystyle x = 0 $

    Hence the equation has a real number solution for all even values of n. And that solution is $\displaystyle x = 0$. 0 is a real number.
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    Senior Member vincisonfire's Avatar
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    Shouldn't it be $\displaystyle

    2(x^2 + 1)^{\frac{n}{2}}\cos{(\frac{n}{x})} = 0
    $ because the angle will be $\displaystyle arctan\frac{1}{x} $
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    Quote Originally Posted by vincisonfire View Post
    Shouldn't it be $\displaystyle

    2(x^2 + 1)^{\frac{n}{2}}\cos{(\frac{n}{x})} = 0
    $ because the angle will be $\displaystyle *arctan\frac{1}{x} $
    Yes you're right, my apologies.

    I did $\displaystyle \frac{\Re{z}}{\Im{z}}$ instead of $\displaystyle \frac{\Im{z}}{\Re{z}}$.
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    Senior Member vincisonfire's Avatar
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    I would thus have $\displaystyle x = \frac{2n}{(2k+1)\pi}, k\in \mathbb Z $ but it does not work.
    And thanks Mush. I guess there is no solution for all n.
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    the solutions are $\displaystyle x=\cot \left(\frac{(2k+1)\pi}{2n} \right), \ 1 \leq k \leq n.$ just put $\displaystyle x=\cot \alpha$ and solve a simple equation.
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    Senior Member vincisonfire's Avatar
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    Mush somewhere you did $\displaystyle *(-1)^{\frac{1}{n}}=\frac{-1}{(-1)^n} $*
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    Quote Originally Posted by vincisonfire View Post
    Mush somewhere you did $\displaystyle *(-1)^{\frac{1}{n}}=\frac{-1}{(-1)^n} $*
    I was wondering about that too.

    Shouldn't it be $\displaystyle \sqrt[n]{-1}$?
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    Senior Member vincisonfire's Avatar
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    To NonCommAlg :
    I don't understand why but as n goes up the solutions are not working???
    e.g. when n = 101 mathematica gives me 2.59828*10^167 + 0. I
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    Senior Member vincisonfire's Avatar
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    Yes it should be $\displaystyle

    \sqrt[n]{-1}
    $. That complicates the things ... but the answer is still good.
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    Quote Originally Posted by vincisonfire View Post
    To NonCommAlg :
    I don't understand why but as n goes up the solutions are not working???
    the solution still work but you won't get any new ones because: $\displaystyle \cot \left(\frac{(2k+1)\pi}{2n} \right)=\cot \left(\frac{(2(k+n)+1)\pi}{2n} \right).$
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    Senior Member vincisonfire's Avatar
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    No I meant $\displaystyle (Cot(\frac{\pi}{202})+i)^{101} + (Cot(\frac{\pi}{202})-i)^{101} = 2.59828*10^{167} + 0. I $
    When n > 10 it begins to have great values.
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    OK this is driving me mad.

    According to my CAS, it reduces to...

    $\displaystyle 2\cos{\left(n\arctan{x} - \frac{n\pi}{2}\right)}(x^2 + 1)^{\frac{n}{2}}$

    and gives

    $\displaystyle x = -\cot{\left(\frac{k\pi}{n} - \frac{\pi}{2n}\right)}$ and $\displaystyle -1 \leq \frac{n + 2k - 1}{n} \leq 1$.
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    Quote Originally Posted by vincisonfire View Post
    No I meant $\displaystyle (Cot(\frac{\pi}{202})+i)^{101} + (Cot(\frac{\pi}{202})-i)^{101} = 2.59828*10^{167} + 0. I $
    When n > 10 it begins to have great values.
    the solution i gave you is exact. you trust the software you're using more than mathematics?!!
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